Before the method: the idea behind it
When the integral is not the area
Last lesson ended on a warning: a definite integral counts region below the x-axis as negative. This lesson is about handling that properly, first for areas between curves and the axis, then between two curves, and finally in motion, where the same distinction becomes the difference between displacement and distance travelled, and where the ClassPad has a trick that makes it almost automatic.
Part one
Signed area
Imagine sweeping from left to right under a curve, adding up area as you go, but with a rule: area below the axis counts as negative.
That’s exactly what a definite integral does. While the curve is above the axis the running total climbs; once it dips below, the total starts falling again. If a curve spends as long below as above, the integral can come out as zero even though there’s plenty of shaded region. So whenever a question asks for area, you need the size of every piece, which means finding where the curve crosses the axis, integrating each piece separately, and adding their absolute values.
Drag b to the right. Rose shading is below the axis and counts negative; teal is above and counts positive. Watch the two readouts drift apart, then meet again.
At b = 3 the integral has climbed all the way back to exactly zero, while the area is 8/3. Same curve, same limits, two different questions. The rule for area: find the x-intercepts inside the interval, split there, and add the sizes of the pieces.
Part two
Area between two curves
The region between two curves is just the region under the top curve with the region under the bottom curve removed.
So integrate the difference, top minus bottom, between the points where the curves meet. This works even when part of the region is below the axis, because subtracting the bottom curve from the top gives the vertical gap between them, and that gap is always positive as long as you have the order right. The only things that can go wrong are getting the intersections wrong, or the curves swapping over partway, in which case you split at the crossing, exactly as in Part one.
Part three
Displacement, distance, and the absolute-value trick
Now put a particle on the axis and let the curve be its velocity.
Velocity is the derivative of position, so integrating velocity gives the change in position: the displacement. And because velocity is negative whenever the particle is moving backwards, the integral automatically subtracts the backwards legs. That’s correct for displacement, which is where you finished relative to where you started. But it’s wrong for distance, which is how far the particle actually moved, backwards legs included. Distance is the area under the velocity graph; displacement is the integral. Part one all over again.
Drag t. From 1 to 3 the velocity is negative: the particle moves back the way it came. Displacement counts that as negative; distance counts it in full.
By t = 4 the particle is only 4/3 from where it started, but it has travelled a full 4 units: out to 4/3, back to 0, out to 4/3 again. On paper you find distance by splitting the integral at every time the velocity is zero. On the ClassPad there is a faster way: integrate the absolute value of the velocity. The bars flip every negative section positive, so a single integral gives the distance directly.
Where this shows up in real life
A trip computer’s two numbers
A car’s trip computer can show “distance travelled” and a GPS can show “distance from home.” After a drive to the shops and back, the first is large and the second is zero. The trip computer is integrating |v|; the GPS is integrating v.
Net versus total, everywhere
Net profit versus total revenue, net water level change versus total rainfall, net charge versus total current: any time a quantity can flow both ways, the integral gives the net and the absolute integral gives the total.
Before you start
The questions to ask yourself
| The question says… | Do this |
|---|---|
| “area” between a curve and the x-axis | Find the x-intercepts in the interval, split there, add absolute values |
| “area enclosed” by two curves | Solve f = g for the intersections, integrate top − bottom between them |
| “displacement”, “change in position” | ∫ v dt, keep the sign |
| “distance travelled” | Split where v = 0 and add sizes, or ∫ |v| dt on the ClassPad |
Next: the working method, with a sketch of the region as the first step of every area question, twenty practice questions, ten harder ones, and the ClassPad guide.